We are interested in increasing the concentration of a 3L methanol-water mixture

Question

We are interested in increasing the concentration of a 3L methanol-water mixture that contains 20% by volume of methanol. The liquid mixture which is initially at 25 degree C is Cola heated to 65 degree C in the still which is connected to other units as shown. Initial pressure is at atmospheric condition. The entire assembly is completely sealed so that no vapors escape. Then vapors are condensed in a condenser by continuously circulating large amount of cold water at 25 degree C. Liquid is collected as a condensate. If 1.0L of liquid with 31.2% (molar basis) of methanol is present as a condensate after 1 h, what will be the condensate composition of the remaining liquid in the still? What volume of liquid will be remaining in the still?

Explanation / Answer

Feed to the column = 3 L of methanol-water mixture. Methanol content in the feed = 3*0.2= 0.6 L

methanol in the liquid condensate = 1*0.312= 0.312 L

methanol remaining in the still = methanol in feed - methanol removed through condenser = 0.6-0.312= 0.288L

water in the feed = 3*0.8=2.4 L, water in the liquid condensate = 1*(1-0.312)= 0.688 L

water in the still = 2.4-0.688=1.712L

composition of liquid in the still ( Vol%): methanol = 0.288/(0.288+1.712) =0.144 or 14.4 % and water = 1.712/((0.288+1.712) =0.856 or 85.6%

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